$h(t) = 3t^{3}+6t^{2}-2t+7+2(f(t))$ $g(x) = 6x$ $f(n) = -2n-g(n)$ $ g(h(-1)) = {?} $
Answer: First, let's solve for the value of the inner function, $h(-1)$ . Then we'll know what to plug into the outer function. $h(-1) = 3(-1)^{3}+6(-1)^{2}+(-2)(-1)+7+2(f(-1))$ To solve for the value of $h$ , we need to solve for the value of $f(-1)$ $f(-1) = (-2)(-1)-g(-1)$ To solve for the value of $f$ , we need to solve for the value of $g(-1)$ $g(-1) = (6)(-1)$ $g(-1) = -6$ That means $f(-1) = (-2)(-1)-(-6)$ $f(-1) = 8$ That means $h(-1) = 3(-1)^{3}+6(-1)^{2}+(-2)(-1)+7+(2)(8)$ $h(-1) = 28$ Now we know that $h(-1) = 28$ . Let's solve for $g(h(-1))$ , which is $g(28)$ $g(28) = (6)(28)$ $g(28) = 168$